Cambridge A Level Mathematics 9709 — 2022 May/June Paper 5 · Variant 1

1 (a) Find the number of different arrangements of the 8 letters in the word DECEIVED in which all three Es are together and the two Ds are together. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the number of different arrangements of the 8 letters in the word DECEIVED in which the three Es are not all together. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) 5! M1 k! where k = 5, 6 or 7 Condone  1 OE 120 A1 2 1(b) [Total no of ways =] 8! 2!3! [= 3360] M1 8! , 1,2 1,3 ! ! a b a b a b    [With 3Es together =] 6! 2! [= 360] M1 6!, 1,2 ! c c  seen in an addition/subtraction [With 3Es not together] = 3360 – 360 M1 8! 6! where , 1,2 & 1,3 ! ! ! d f e d e f    3000 A1 4

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2 There are 6 men and 8 women in a Book Club. The committee of the club consists of five of its members. Mr Lan and Mrs Lan are members of the club. (a) In how many different ways can the committee be selected if exactly one of Mr Lan and Mrs Lan must be on the committee? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) In how many different ways can the committee be selected if Mrs Lan must be on the committee and there must be more women than men on the committee? [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) M1 990 A1 Alternative method for question 2(a) [total – both on – neither on] 14C5 – ( 12C3 + 12C5) = [2002 – 220 – 792] M1 kC5 – (aC3 + aC5) a = 12, 13 and k = 13, 14 990 A1 2 2(b) [Mrs Lan plus] 2W 2M 7C2  6C2 = 315 3W 1M 7C3  6C1 = 210 4W 7C4 = 35 M1 7Cr  6C4-r for r = 2, 3 or 4 B1 Outcome for one identifiable scenario correct, accept unevaluated M1 Add outcomes for 3 identifiable correct scenarios Note: if scenarios not labelled, they may be identified by seeing7Cr  6Cs r + s = 4 to imply r women and s men for both B & M marks only [Total =] 560 A1 4

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3 The times taken to travel to college by 2500 students are summarised in the table. Time taken (t minutes) 0 ≤t < 20 20 ≤t < 30 30 ≤t < 40 40 ≤t < 60 60 ≤t < 90 Frequency 440 720 920 300 120 (a) Draw a histogram to represent this information. [4] From the data, the estimate of the mean value of t is 31.44. (b) Calculate an estimate of the standard deviation of the times taken to travel to college. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) In which class interval does the upper quartile lie? [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It was later discovered that the times taken to travel to college by two students were incorrectly recorded. One student’s time was recorded as 15 instead of 5 and the other’s time was recorded as 65 instead of 75. (d) Without doing any further calculations, state with a reason whether the estimate of the standard deviation in part (b) would be increased, decreased or stay the same. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Class width 20 10 10 20 30 Frequency density 22 72 92 15 4 (Frequency ÷ class width, e.g. 440 440 440 , 20 19.5 20.5       condone Accept unsimplified, may be read from graph using their scale A1 All heights correct on graph NOT FT B1 Bar ends at [0,] 20, 30, 40, 60, 90 at axis with a horizontal linear scale with at least 3 values indicated. 0 ⩽ horizontal scale ⩽ 90 B1 Axes labelled frequency density (fd), time (t) and minutes (mins) or in a title. Linear vertical scale, with at least 3 values indicated 0 ⩽ vertical axes ⩽ 92 (condone 90 used). 4 Question Answer Marks Guidance 3(b) Midpoints 10 25 35 50 75 B1 At least 4 correct midpoints seen [Mean = 31.44 given] [Variance = 2 2 2 2 2 2 440 10 720 25 920 35 300 50 120 75 31.44 2500           ] = 2 44000 450000 1127000 750000 675000 31.44 2500      [= 2 3046000 31.44 229.9264 2500   ] Or Variance = 2 2 2 2 2 440(10 31.44) 720(25 31.44) 920(35 31.44) 300(50 31.44) 120(75 31.44) 2500          202256 29860 11659 103342 227697 574814 229.9264 2500 2500        M1 Correct formula for variance or standard deviation (− mean2 included with their midpoints (not upper bound, lower bound, class width, frequency density, frequency or cumulative frequency) and their ∑f if calculated. Condone 1 data error. Standard deviation = 15.2 A1 WWW, allow 15.16[3…] 3 3(c) 30‒40 B1 1 3(d) Stays the same, data still in same intervals B1 Frequencies unchanged 1

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5 The lengths, in cm, of the leaves of a particular type are modelled by the distribution N 5.2, 1.52 . (a) Find the probability that a randomly chosen leaf of this type has length less than 6cm. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The lengths of the leaves of another type are also modelled by a normal distribution. A scientist measures the lengths of a random sample of 500 leaves of this type and finds that 46 are less than 3cm long and 95 are more than 8cm long. (b) Find estimates for the mean and standard deviation of the lengths of leaves of this type. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) In a random sample of 2000 leaves of this second type, how many would the scientist expect to find with lengths more than 1 standard deviation from the mean? [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) P(X < 6) = P(Z < 6 5.2) P( 0.5333) 1.5 Z   M1 6, 5.2, 1.5 substituted into ± standardisation formula, condone 1.52, continuity correction 0.5  0.703 A1 2 5(b) 1 3 1.329      z 2 8 0.878      z B1 1.328 < z1 ⩽ 1.329 or −1.329 ⩽ z1 < −1.328 B1 0.877 < z2 ⩽ 0.878 or −0.878 ⩽ z2 < −0.877 Solve to find at least one unknown: 3 1.329     8 0.878     M1 Use of the ± standardisation formula once with μ, σ, a z-value (not 0.8179, 0.7910, 0.5367, 0.5753, 0.19, 0.092 etc.) and 3 or 8, condone continuity correction but not σ2 or √σ M1 Use either the elimination method or the substitution method to solve their two equations in μ and σ 2.27, 6.01     A1 2.26 ⩽ σ ⩽ 2.27, 6.01 ⩽ μ ⩽ 6.02 5 Question Answer Marks Guidance 5(c) [P(Z < −1) + P(Z >1)] Φ(1) – Φ(−1) = = 2 – 2  Φ 1 = 2 – 2  0.8413 M1 Identify 1 and –1 as the appropriate z-values. M1 Calculating the appropriate area from stated phis of z-values which must be ± the same number 0.3174 A1 Accept AWRT 0.317 Number of leaves: 2000  0.3174 = 634.8 so 634 or 635 B1 FT FT their 4 s.f. (or better) probability, final answer must be positive integer no approximation or rounding stated 4

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6 Janice is playing a computer game. She has to complete level 1 and level 2 to finish the game. She is allowed at most two attempts at any level. • For level 1, the probability that Janice completes it at the first attempt is 0.6. If she fails at her first attempt, the probability that she completes it at the second attempt is 0.3. • If Janice completes level 1, she immediately moves on to level 2. • For level 2, the probability that Janice completes it at the first attempt is 0.4. If she fails at her first attempt, the probability that she completes it at the second attempt is 0.2. (a) Show that the probability that Janice moves on to level 2 is 0.72. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that Janice finishes the game. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the probability that Janice fails exactly one attempt, given that she finishes the game. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) 0.6 0.4 0.3 0.72 or 1 – 0.4 x 0.7 = 0.72 1 6(b)   0.72 0.4 0.6 0.2    M1 0.72  u, 0 < u < 1 M1 v  (0.4 + 0.6  0.2), or v  (1 – 0.6  0.8) 0 < v ⩽ 1 no additional terms SC B1 for 0.72  (0.4 + 0.12) or 0.72  (1 – 0.48) 0.3744 A1 WWW. Condone 0.374. SC B1 for 0.3744 only 3 Alternative method for question 6(b) [p(P1P2) + p(F1P1P2) + p(P1F2P2) + p(F1P1F2P2)] = 0.6  0.4 + 0.4  0.3  0.4 + 0.6  0.6  0.2 + 0.4  0.3  0.6  0.2 M1 Any two terms unsimplified and correct M1 Summing 4 appropriate scenarios by listing or on a tree diagram SC B1 for 0.24 + 0.048 + 0.072 + 0.0144 0.3744 A1 WWW. Condone 0.374. SC B1 for 0.3744 only 3 Question Answer Marks Guidance 6(c) (fails first or second level finishes game) (fails first or second level | finishes game)  P P their(b) Numerator = P(S SF) + P(FS S) = 0.6  0.6  0.2 + 0.4  0.3  0.4 = 0.072 + 0.048 = 0.12 Required probability = 0.12 their(b) M1 Either 0.6  0.6  0.2 or 0.4  0.3  0.4 seen Condone 0.072 or 0.048 if seen in (b) A1 Both correct accept unsimplified expression. No additional terms M1 sum of two 3-term probabilities as numerator or correct Their their (b) 0.321 or 25 78 A1 0.3205 < p ⩽ 0.321 4

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